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Chapter: Pythagoras Theorem· Mathematics Part 2 (Geometry)

Asked in: March 2025 board paperComplete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.Given: \(\square PQRS\) is a rhombus. Diagonals PR and SQ intersect each other…

Asked in: March 2025 board paper

Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.

Given: \(\square PQRS\) is a rhombus. Diagonals PR and SQ intersect each other at point T. To prove: \(PS^2 + SR^2 + QR^2 + PQ^2 = PR^2 + QS^2\).

Activity: Diagonals of a rhombus bisect each other. In \(\Delta PQS\), PT is the median and in \(\Delta QRS\), RT is the median. \(\therefore\) by Apollonius theorem, \(PQ^2 + PS^2 = \text{____} + 2QT^2\) ...... (I); \(QR^2 + SR^2 = \text{____} + 2QT^2\) ...... (II). Adding (I) and (II), \(PQ^2 + PS^2 + QR^2 + SR^2 = 2(PT^2 + \text{____}) + 4QT^2 = 2(PT^2 + \text{____}) + 4QT^2\) (RT = PT) \(= 4PT^2 + 4QT^2 = (\text{____})^2 + (2QT)^2\) \(\therefore PQ^2 + PS^2 + QR^2 + SR^2 = PR^2 + \text{____}\).

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Answer: Blanks in order: \(2PT^2\), \(2RT^2\), \(RT^2\), \(PT^2\), \(2PT\), \(QS^2\)

Diagonals of a rhombus bisect each other, so T is the midpoint of PR and of QS: PT = RT and QT = ST.

In \(\Delta PQS\), PT is a median: \(PQ^2 + PS^2 = \boxed{2PT^2} + 2QT^2\) ...... (I) (Apollonius theorem)

In \(\Delta QRS\), RT is a median: \(QR^2 + SR^2 = \boxed{2RT^2} + 2QT^2\) ...... (II)

Adding: \(PQ^2 + PS^2 + QR^2 + SR^2 = 2(PT^2 + \boxed{RT^2}) + 4QT^2 = 2(PT^2 + \boxed{PT^2}) + 4QT^2\) (RT = PT)

\(= 4PT^2 + 4QT^2 = (\boxed{2PT})^2 + (2QT)^2\)

Since 2PT = PR and 2QT = QS, \(PQ^2 + PS^2 + QR^2 + SR^2 = PR^2 + \boxed{QS^2}\). Hence proved.

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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